Question

0.472 M sulfuric acid was used to titrate a 0.388 M barium hydroxide solution. The initial volume reading on the acid buret was 34.42 mL and the final reading was 41.90 mL. If the initial reading on the base buret was 63.25 mL, what was the final volume reading on the base

Answer 1

Answer: The final volume reading on the base is 72.35 mL

Explanation:

To calculate the volume of base, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2SO_4$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is $Ba(OH)_2$

We are given:

$n_1=2\\M_1=0.472M\\V_1=[41.90-34.42]=7.48mL\\n_2=2\\M_2=0.388M\\V_2=?mL$

Putting values in above equation, we get:

$2\times 0.472\times 7.48=2\times 0.388\times V_2\\\\V_2=\frac{2\times 0.472\times 7.48}{2\times 0.388}=9.10mL$

Final volume of barium hydroxide solution = Initial volume + Volume needed to neutralize = [63.25 + 9.10] = 72.35 mL

Hence, the final volume reading on the base is 72.35 mL